<span>let x=gallons of current mixture to be drained
and replaced with pure antifreeze.
4-x=gallons of current mixture remaining in the car.</span>
<span>
0.15(4-x)+1.00x=0.50 x 4
0.6-.15x+x=2
0.85x=1.4
x=1.4/0.85 =1.65 gal
Thus, 1.65 gallons of current mixture to be drained and replaced with pure
antifreeze.</span>
The answer to this question would be: <span> 10 °K
Kelvin and Celcius scales are different by 273</span> degrees but their ratio is the same. One degree in Kelvin is equal to one degree in Celcius. That mean, 10 °C change in Celcius would be same as <span> 10 °K changes in Kelvin too. </span>
Following are important constant that used in present calculations
Heat of fusion of H2O = 334 J/g
<span>Heat of vaporization of H2O = 2257 J/g </span>
<span>Heat capacity of H2O = 4.18 J/gK
</span>
Now, energy required for melting of ICE = <span> 334 X 5.25 = 1753.5 J .......(1)
Energy required for raising </span><span>the temperature water from 0 oC to 100 oC = 4.18 X 5.25 X 100 = 2195.18 J .............. (2)
</span>Lastly, energy required for boiling water = <span> 2257X 5.25 = 11849.25 J ......(3)
</span><span>
Thus, total heat energy required for entire process = (1) + (2) + (3)
= 1753.5 + 2195.18 + 11849.25
= </span><span>15797.93 J
</span><span> = 15.8 kJ
</span><span>Thus, 15797.93 J of energy is needed to boil 5.25 grams of ice.</span>
